compactness

• May 3rd 2009, 04:25 AM
jin_nzzang
compactness
The Hilbert cube Q is the subset of l given by {a : NR││a(i)│2-i}.
Show that Q is a compact subspace of l∞ (with the norm a=supnN│a(n)│).

• May 4th 2009, 10:40 AM
Opalg
Quote:

Originally Posted by jin_nzzang
The Hilbert cube Q is the subset of l given by {a : NR││a(i)│2-i}.
Show that Q is a compact subspace of l∞ (with the norm a=sup_{nN}│a(n)│).

Let $\mathcal B$ denote the unit ball of $l^\infty$. Then $\mathcal B$ consists of all sequences $(x_n)$ of real numbers such that $x_n\in[-1,1]$ for all n. Give $\mathcal B$ the topology that it has as the product of an infinite number of copies of [–1,1]. In this topology, a sequence $((x^{(k)}))$ converges to $((x^{(0)}))$ if it converges coordinatewise. That is to say, if $x^{(k)}_n\to x^{(0)}_n$ for each coordinate n.

The map $f:(x_n)\mapsto(2^{-n}x_n)$ takes $\mathcal B$ onto Q. The idea of the proof is to show that f is continuous from $\mathcal B$ (with the above product topology) to Q (with the topology from the $l^\infty$ norm). By Tychonoff's theorem, any product of compact spaces is compact. So $\mathcal B$ is compact, and hence Q, as a continuous image of a compact space, is also compact.

To see that f is continuous, suppose that $x^{(k)}\to x^{(0)}$ in $\mathcal B$, and let $\varepsilon>0$. Choose N so that $2^{-N}<\varepsilon/2$. Then $|(f(x^{(k)}) - f(x^{(0)}))_n| = 2^{-n}|x^{(k)}_n - x^{(0)}_n| < \varepsilon$ for all n>N. But (by the definition of the product topology in $\mathcal B$) there exists K such that $|x^{(k)}_n - x^{(0)}_n| < \varepsilon$ for all k>K and all n with 1≤n≤N. Therefore $\|f(x^{(k)}) - f(x^{(0)})\|_\infty \leqslant \varepsilon$, so that $f(x^{(k)})\to f(x^{(0)})$ in Q.

[That is essentially the argument given here to show that the Hilbert cube is compact for the topology that comes from the $l^2$-norm, except that the Wikipedia page completely ducks the central part of the proof, which is to show that the map f is continuous.]